Subsection 7.2.1 Iterating Differentiation
Starting with a differentiable function \(f(x)\text{,}\) I used the derivative to get a whole new function \(f^\prime(x)\) which measured the slope of the previous function. This solves the velocity problem: if \(p(t)\) is position of an object, then \(p^\prime(t)\) is the velocity of the object.
I can continue this process. If \(f^\prime(x)\) is still a differentiable function itself, I can take another derivative to get \(f^{\prime \prime}(x)\text{.}\) This is called the second derivative. The process is exactly the same: I use the same limit definition and the same differentiation rules to find this second derivative. \(f^{\prime \prime}(t)\) is Newton’s notation for the second derivative; the Leibniz notation is \(\frac{d^2 f}{dx^2}\text{.}\) If \(p(t)\) was a position function and \(p^\prime(t)\) was its velocity (the rate of change of position), the \(p^{\prime \prime}(t)\) would be the rate of change of velocity. That woud be acceleration. Acceleration is the second derivative of position.
This has important implications for physics. Newton’s first law of motion, \(F = ma\text{,}\) says that force equals mass times acceleration. If force depends on position (as it often does), then I can write Newton’s first law as a differential equation.
\begin{equation*}
F(p) = m \frac{d^2 p}{dt^2}
\end{equation*}
As with most of the fundamentals of physics, Newtonian motion is determined by the solution of a differential equation. I can also give a more specific example. On the surface of the earth, it is assumed that the acceleration due to gravity is constant at roughly \(9.8 m/s^2\text{.}\) I can observe that the flight of projectiles is roughly parabolic, that is, height is described by a parabola \(h(t) = at^2 + bt + c\text{.}\) If I differentiate once, vertical velocity is \(h^\prime(t) = 2at + b\text{.}\) If I differentiate twice, vertical acceleration is \(h^{\prime
\prime}(t) = 2a\text{,}\) which is constant. So parabolic paths fit with the notion of a constant acceleration due to gravity. Even more, I know that the leading coefficient of those quadratics should be roughly \(a = 4.9\) to get the constant acceleration \(2a = 9.8m/s^2\text{.}\)
Another specific example is Hooke’s law. This law describes the force caused by a spring; the law states that the force is (negatively) proportional to the distance from equalibrium. If equalibrium is at \(p=0\text{,}\) then \(F = -kp\text{.}\) Using Newton’s first law with this force produces a differential equation.
\begin{equation*}
-kp = m \frac{d^2 p}{dt^2}
\end{equation*}
The behaviour of an object on a (perfect, frictionless) spring is determined by this differential equation. What is that behaviour? I simply have to figure out which function \(p(t)\) matches this differential equation. In this case, there are two solutions.
\begin{align*}
p(t) \amp = \sin \left( \sqrt{\frac{m}{k}} t \right) \amp
\text{or} \amp \amp p(t) \amp= \cos \left( \sqrt{\frac{m}{k}}
t \right)
\end{align*}
Since the sinusoidal functions solve the differential equation, I can conclude that an object on a spring should act with a sinusoidal behaviour.
